help me with my math homework, please?

Question

Math homework help, please?????
the denominator of a fraction is a two-digit number whose digits add up to ten. If the digits are reversed, the new number is the numerator. if 41 is subtracted from the numerator, the value of the resulting fraction is 1/2. find the original fraction

Answer 1

let the at units place of the denominator be x.Therefore,the digit at the ten's pkace is 10-x
hence,the denominator=10(10-x)+x
=100-10x+x=100-9x
When the digits of the denominator is intrchange,it is equal to numerator.
Therefore,the numerator=10*x+(10-x)
=10x+10-x=9x+10
Therefore,the fraction is (9x+10)/(100-9x)
by the problem,
(9x+10-41)/(100-9x)=1/2
=>(9x-31)/(100-9x)=1/2
=>18x-62=100-9x
=>18x+9x=100+62
=>27x=162
=>x=162/27=6
Therefore,the original fraction is
(9*6+10)/100-9*6
=( 54+10)/100-54)
=64/46

Answer 2

Let N be the numberator and D be the denominator
D = 10i+j where i,j are integers
10j+i = N
(N-41)/D = 1/2
Now sub in to the above eqn:
(10j+i - 41)/(10i+j) = 1/2
2(10j+i-41) = 10i+j
20j+2i-82 = 10i+j
i+j=10
i = 10-j
20j+2(10-j)-82 = 10i+j = 10(10-j) + j = 100 - 9j
20j + 20 - 2j - 82 = 100 - 9j
27j = 162
j = 162/27 = 6
i = 10-j = 10-6 = 4
N = 10j+i = 64
D = 10i+j = 46
(N-41)/D = (64-41)/46 = 23/46 = 1/2 as required

Answer 3

subtracting 41 cuts the nimber in 1/2, therefore the numerator must be 41*2=82 & the denominator is 28

82/28 is the original fraction.

Answer 4

1=9=10
2+8=10
3+7=10
4+6=10
5+5=10

Answer 5

64/46

Answer 6

64/46